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I am a little confused with some aspects of electrical power and electrical work. The example is connected to a thermodynamic problem:

Consider an adiabatic vessel filled with an arbitrary fluid. Inside the fluid, there is a electric heating coil connected to a power supply at the outside of the vessel. Heat conduction and convection limitations should be neglected.

Consider an adiabatic vessel filled with an arbitrary fluid. Inside the fluid, there is a electric heating coil connected to a power supply at the outside of the vessel. Heat conduction and convection limitations should be neglected.

From the first law of thermodynamics:

$dU = dQ + dW_{el}$ and $dQ = 0$, since the vessel is adiabatic.

Now I am looking for $\frac{dU}{dt}$ and the following equation should be valid: $\frac{dU}{dt}=\frac{dW_{el}}{dt}$.

Since, $W_{el} = E(t)\cdot C(t)$, where $E$ is the voltage and $C$ the charge ($E$ and $C$ to not conflict the thermodynamic nomenclature). Therefore, I thought $\frac{W_{el}}{dt} = \frac{dE(t)}{dt}\cdot C(t) + \frac{dC(t)}{dt}\cdot E(t)$ with $\frac{dC(t)}{dt} = I(t)$.

Unfortunately, I found the following $\frac{dU}{dt}=E(t)\cdot I(t)$, which differ in the term $\frac{E(t)}{dt}\cdot C(t)$. I don't understand where my mistake is, can somebody please clarify?

I am a little confused with some aspects of electrical power and electrical work. The example is connected to a thermodynamic problem:

Consider an adiabatic vessel filled with an arbitrary fluid. Inside the fluid, there is a electric heating coil connected to a power supply at the outside of the vessel. Heat conduction and convection limitations should be neglected.

From the first law of thermodynamics:

$dU = dQ + dW_{el}$ and $dQ = 0$, since the vessel is adiabatic.

Now I am looking for $\frac{dU}{dt}$ and the following equation should be valid: $\frac{dU}{dt}=\frac{dW_{el}}{dt}$.

Since, $W_{el} = E(t)\cdot C(t)$, where $E$ is the voltage and $C$ the charge ($E$ and $C$ to not conflict the thermodynamic nomenclature). Therefore, I thought $\frac{W_{el}}{dt} = \frac{dE(t)}{dt}\cdot C(t) + \frac{dC(t)}{dt}\cdot E(t)$ with $\frac{dC(t)}{dt} = I(t)$.

Unfortunately, I found the following $\frac{dU}{dt}=E(t)\cdot I(t)$, which differ in the term $\frac{E(t)}{dt}\cdot C(t)$. I don't understand where my mistake is, can somebody please clarify?

I am a little confused with some aspects of electrical power and electrical work. The example is connected to a thermodynamic problem:

Consider an adiabatic vessel filled with an arbitrary fluid. Inside the fluid, there is a electric heating coil connected to a power supply at the outside of the vessel. Heat conduction and convection limitations should be neglected.

From the first law of thermodynamics:

$dU = dQ + dW_{el}$ and $dQ = 0$, since the vessel is adiabatic.

Now I am looking for $\frac{dU}{dt}$ and the following equation should be valid: $\frac{dU}{dt}=\frac{dW_{el}}{dt}$.

Since, $W_{el} = E(t)\cdot C(t)$, where $E$ is the voltage and $C$ the charge ($E$ and $C$ to not conflict the thermodynamic nomenclature). Therefore, I thought $\frac{W_{el}}{dt} = \frac{dE(t)}{dt}\cdot C(t) + \frac{dC(t)}{dt}\cdot E(t)$ with $\frac{dC(t)}{dt} = I(t)$.

Unfortunately, I found the following $\frac{dU}{dt}=E(t)\cdot I(t)$, which differ in the term $\frac{E(t)}{dt}\cdot C(t)$. I don't understand where my mistake is, can somebody please clarify?

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derive Derive electrical power from electrical work

I am a little confused with some aspects of electrical power and electrical work. The example is connected to a thermodynamic problem:

Consider an adiabatic vessel filled with an arbitrary fluid. Inside the fluid, there is a electric heating coil connected to a power supply at the outside of the vessel. Heat conduction and convection limitations should be neglected.

From the first law of thermodynamics:

$dU = dQ + dW_{el}$ and $dQ = 0$, since the vessel is adiabatic.

Now I am looking for $\frac{dU}{dt}$ and the following equation should be valid: $\frac{dU}{dt}=\frac{dW_{el}}{dt}$.

Since, $W_{el} = E(t)\cdot C(t)$, where E$E$ is the voltage and C$C$ the charge (E$E$ and C$C$ to not conflict the thermodynamic nomenclature). Therefore, I thought $\frac{W_{el}}{dt} = \frac{dE(t)}{dt}\cdot C(t) + \frac{dC(t)}{dt}\cdot E(t)$ with $\frac{dC(t)}{dt} = I(t)$.

Unfortunately, I found the following $\frac{dU}{dt}=E(t)\cdot I(t)$, which differ in the term $\frac{E(t)}{dt}\cdot C(t)$. I don't understand where my mistake is, can somebody please clarify?

Thanks!

derive electrical power from electrical work

I am a little confused with some aspects of electrical power and electrical work. The example is connected to a thermodynamic problem:

Consider an adiabatic vessel filled with an arbitrary fluid. Inside the fluid, there is a electric heating coil connected to a power supply at the outside of the vessel. Heat conduction and convection limitations should be neglected.

From the first law of thermodynamics:

$dU = dQ + dW_{el}$ and $dQ = 0$, since the vessel is adiabatic.

Now I am looking for $\frac{dU}{dt}$ and the following equation should be valid: $\frac{dU}{dt}=\frac{dW_{el}}{dt}$.

Since, $W_{el} = E(t)\cdot C(t)$, where E is the voltage and C the charge (E and C to not conflict the thermodynamic nomenclature). Therefore, I thought $\frac{W_{el}}{dt} = \frac{dE(t)}{dt}\cdot C(t) + \frac{dC(t)}{dt}\cdot E(t)$ with $\frac{dC(t)}{dt} = I(t)$.

Unfortunately, I found the following $\frac{dU}{dt}=E(t)\cdot I(t)$, which differ in the term $\frac{E(t)}{dt}\cdot C(t)$. I don't understand where my mistake is, can somebody please clarify?

Thanks!

Derive electrical power from electrical work

I am a little confused with some aspects of electrical power and electrical work. The example is connected to a thermodynamic problem:

Consider an adiabatic vessel filled with an arbitrary fluid. Inside the fluid, there is a electric heating coil connected to a power supply at the outside of the vessel. Heat conduction and convection limitations should be neglected.

From the first law of thermodynamics:

$dU = dQ + dW_{el}$ and $dQ = 0$, since the vessel is adiabatic.

Now I am looking for $\frac{dU}{dt}$ and the following equation should be valid: $\frac{dU}{dt}=\frac{dW_{el}}{dt}$.

Since, $W_{el} = E(t)\cdot C(t)$, where $E$ is the voltage and $C$ the charge ($E$ and $C$ to not conflict the thermodynamic nomenclature). Therefore, I thought $\frac{W_{el}}{dt} = \frac{dE(t)}{dt}\cdot C(t) + \frac{dC(t)}{dt}\cdot E(t)$ with $\frac{dC(t)}{dt} = I(t)$.

Unfortunately, I found the following $\frac{dU}{dt}=E(t)\cdot I(t)$, which differ in the term $\frac{E(t)}{dt}\cdot C(t)$. I don't understand where my mistake is, can somebody please clarify?

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derive electrical power from electrical work

I am a little confused with some aspects of electrical power and electrical work. The example is connected to a thermodynamic problem:

Consider an adiabatic vessel filled with an arbitrary fluid. Inside the fluid, there is a electric heating coil connected to a power supply at the outside of the vessel. Heat conduction and convection limitations should be neglected.

From the first law of thermodynamics:

$dU = dQ + dW_{el}$ and $dQ = 0$, since the vessel is adiabatic.

Now I am looking for $\frac{dU}{dt}$ and the following equation should be valid: $\frac{dU}{dt}=\frac{dW_{el}}{dt}$.

Since, $W_{el} = E(t)\cdot C(t)$, where E is the voltage and C the charge (E and C to not conflict the thermodynamic nomenclature). Therefore, I thought $\frac{W_{el}}{dt} = \frac{dE(t)}{dt}\cdot C(t) + \frac{dC(t)}{dt}\cdot E(t)$ with $\frac{dC(t)}{dt} = I(t)$.

Unfortunately, I found the following $\frac{dU}{dt}=E(t)\cdot I(t)$, which differ in the term $\frac{E(t)}{dt}\cdot C(t)$. I don't understand where my mistake is, can somebody please clarify?

Thanks!